Two-Dimensional Momentum: Vector Conservation and Glancing Collisions

1) Momentum in Two Dimensions: Conservation as a Vector Rule

In two-dimensional collisions, momentum is not a single number; it is a vector. The conservation law is still one statement, but it contains two independent component equations:

\vec p_{\text{before}} = \vec p_{\text{after}}

For a system of objects (often two colliding bodies) with negligible external impulse during the short collision, the total momentum vector is conserved. In component form:

• \sum p_x\,\text{(before)} = \sum p_x\,\text{(after)}
• \sum p_y\,\text{(before)} = \sum p_y\,\text{(after)}

This “split into x and y” is not a trick; it is exactly what vector equality means. The key skill is choosing axes that make the algebra simpler (often align x with an incoming velocity).

Component setup template

For each object i, write its momentum components using its speed and direction:

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• p_{ix} = m_i v_i \cos\theta_i
• p_{iy} = m_i v_i \sin\theta_i

Then sum over objects before and after.

Diagram-first workflow (recommended)

1. Sketch the situation: draw incoming velocity vectors and outgoing velocity vectors. Mark angles relative to a chosen x-axis.

2. Choose axes: often x along the initial motion of one object; y perpendicular.

3. Write momentum conservation in x and y: two equations.

4. Count unknowns: if you have more unknowns than equations, you need additional information (like “elastic” or a measured angle).

5. Validate: check units, angle reasonableness, and speed bounds.

Quick example: “initially at rest” target makes y-momentum start at zero

If object 2 is initially at rest, then initially p_{2x}=0 and p_{2y}=0. If object 1 initially moves along +x, then initially total y-momentum is zero. Therefore after the collision:

\sum p_y\,\text{(after)} = 0

This forces the outgoing y-components to cancel (equal magnitude, opposite sign) even if both objects move at angles.

2) Glancing Collisions: Unknown Angles and What Extra Information You Need

A glancing collision is one where the objects do not move along a single line after impact; they scatter at angles. In 2D, momentum conservation alone gives only two equations (x and y), so you must be careful about how many unknowns you are trying to solve for.

Typical glancing collision scenario

Object 1 (mass m) moves with known speed v along +x. Object 2 (mass M) is initially at rest. After collision, object 1 leaves with speed v_1' at angle \theta, and object 2 leaves with speed v_2' at angle \phi.

Unknowns could be: v_1', v_2', \theta, \phi (four unknowns). Momentum gives two equations, so you need two more pieces of information.

Set up the momentum equations

With x along the initial motion of object 1:

x-direction

m v = m v_1'\cos\theta + M v_2'\cos\phi

y-direction

0 = m v_1'\sin\theta - M v_2'\sin\phi

(The minus sign assumes the objects scatter to opposite sides of the x-axis; your diagram determines the signs.)

What additional information can close the system?

• Elastic collision constraint: kinetic energy is conserved during the collision. This adds one equation:

  \tfrac12 m v^2 = \tfrac12 m {v_1'}^2 + \tfrac12 M {v_2'}^2

  Now you have three equations.

• A measured angle (e.g., \theta known from video analysis) or a measured speed (e.g., v_1' from a sensor). Each measurement adds an equation.

• Special geometry (common in idealized billiards): for equal masses with one initially at rest and an elastic collision, the outgoing velocities are perpendicular. This provides a geometric relation between angles (an extra constraint).

Important: “Elastic” alone is not always enough to solve everything unless you also know at least one angle or have a special mass relationship. Always count unknowns.

Worked setup: elastic glancing collision with one measured angle

Suppose m, M, and initial v are known, and after collision you measure \theta for object 1. Unknowns: v_1', v_2', \phi (three unknowns). Use:

1. Momentum y to relate v_2' and \phi to v_1':

   M v_2'\sin\phi = m v_1'\sin\theta

2. Momentum x to connect the x-components:

   M v_2'\cos\phi = m v - m v_1'\cos\theta

3. Divide the two to eliminate v_2' and solve for \phi in terms of v_1':

   \tan\phi = \dfrac{m v_1'\sin\theta}{m v - m v_1'\cos\theta}

4. Use energy conservation to solve for v_1' and v_2':

   m v^2 = m {v_1'}^2 + M {v_2'}^2

   Combine with either momentum equation to eliminate v_2'.

This is the standard pattern: momentum gives direction-coupling; energy (if elastic) gives speed-coupling.

Special case: equal masses, target initially at rest, elastic

Let m=M, object 2 initially at rest, and the collision is elastic. Then the outgoing velocity vectors are perpendicular:

\vec v_1' \cdot \vec v_2' = 0

This implies:

\theta + \phi = 90^\circ

Why this matters: if you measure one scattering angle, you immediately know the other, and momentum can then determine the speeds.

Validation steps for glancing-collision answers

• Angle reasonableness: if initial y-momentum is zero, outgoing y-momenta must cancel. Your angles and speeds should reflect opposite y-components.

• Speed bounds: in an elastic collision with one object initially at rest, neither object should emerge with a speed that violates energy conservation. A quick check is that total kinetic energy after equals before.

• Unit consistency: momentum terms must be in kg·m/s. If you accidentally mix degrees/radians inside trig functions in a calculator, the numbers will be wrong even though units look fine.

3) Applications and Combined Methods

A) Billiards-like interactions (idealized)

In an idealized billiards collision between two balls on a level table, external impulses during the short impact are small compared with the contact impulse, so momentum in the plane of the table is approximately conserved during the collision.

Diagram-first approach:

1. Draw the cue ball incoming velocity along +x.

2. Draw the line of centers at impact (the direction along which the contact force acts). This line is crucial because it often determines how momentum is exchanged.

3. Resolve velocities into components parallel and perpendicular to the line of centers. In many ideal models (smooth balls), the impulse acts mainly along the line of centers, so the perpendicular component of each ball’s velocity may be unchanged through the collision.

Practical takeaway: choosing axes aligned with the line of centers can simplify the math even more than choosing axes aligned with the incoming velocity.

B) Spacecraft docking (highly idealized impulse model)

Consider two spacecraft modules that make contact and latch. During the short docking impulse, treat external forces as negligible compared with the contact forces between modules (an approximation that can be reasonable over a brief time). Then total momentum of the two-module system is conserved.

In 2D, if module A approaches with velocity \vec v_A and module B has velocity \vec v_B, and they stick together (perfectly inelastic docking), the final combined velocity is:

\vec v_f = \dfrac{m_A\vec v_A + m_B\vec v_B}{m_A + m_B}

This is a vector equation: compute x and y components separately.

Validation checks:

• Direction: \vec v_f must lie between the directions of \vec v_A and \vec v_B (a mass-weighted average).

• Speed bound: |\vec v_f| cannot exceed the larger of the two initial speeds if the other module is not moving faster in the same direction; it is an average, not an amplification.

C) Ballistic pendulum (momentum + energy in one problem)

The ballistic pendulum is a classic example where you combine momentum conservation during a short collision with energy analysis during a slower swing. In 2D, you may also need vector components if the projectile hits off-center or the pendulum is allowed to swing in a plane with a defined direction.

Idealized straight-on version (planar, no sideways motion):

• A projectile of mass m with speed v embeds in a block of mass M initially at rest.

• Immediately after impact, the combined mass m+M moves with speed V.

• Then it swings upward to a maximum height change h.

Step 1: collision (momentum)

m v = (m+M) V

Step 2: swing (energy)

\tfrac12 (m+M) V^2 = (m+M) g h

From the energy step: V = \sqrt{2gh}. Substitute into momentum to find the projectile speed:

v = \dfrac{m+M}{m}\sqrt{2gh}

Ballistic pendulum with sideways (2D) motion: what changes?

If the projectile arrives at an angle in the horizontal plane, the collision momentum equation becomes vector:

m\vec v = (m+M)\vec V

So:

• V_x = \dfrac{m}{m+M} v_x
• V_y = \dfrac{m}{m+M} v_y

The swing height depends on the magnitude of the post-collision speed:

|\vec V| = \sqrt{V_x^2 + V_y^2}

Then use \tfrac12 (m+M)|\vec V|^2 = (m+M)gh. This shows a common pattern in 2D problems: momentum is handled component-by-component, while energy often uses speed magnitude.

Problem-Solving Toolkit: Tables, Diagrams, and Checks

Momentum component table (fill-in format)

Then enforce:

• \sum p_x\text{ before} = \sum p_x\text{ after}
• \sum p_y\text{ before} = \sum p_y\text{ after}

Diagram checklist (what to label)

• Masses m_1, m_2

• Known initial velocities with arrows and numeric values

• Unknown final velocities with arrows and labeled angles

• Chosen axes (x, y) clearly drawn

• Signs of y-components (above/below x-axis)

Common validation tests

• Component conservation test: plug your final values back into both x and y momentum equations; both must match the initial totals.

• Energy consistency (if elastic): compute total kinetic energy before and after; they must match within rounding.

• Extreme-case sanity: if an angle goes to 0, your equations should reduce to a mostly 1D-like outcome in x with near-zero y-components.

• Units: momentum terms in kg·m/s, energy terms in J, angles handled consistently in your calculator.